Exercise 4 Clustering and classification.

#Access the MASS package.
library(MASS)
#Load the data.
data("Boston")
#Explore the dataset.
str(Boston)
## 'data.frame':    506 obs. of  14 variables:
##  $ crim   : num  0.00632 0.02731 0.02729 0.03237 0.06905 ...
##  $ zn     : num  18 0 0 0 0 0 12.5 12.5 12.5 12.5 ...
##  $ indus  : num  2.31 7.07 7.07 2.18 2.18 2.18 7.87 7.87 7.87 7.87 ...
##  $ chas   : int  0 0 0 0 0 0 0 0 0 0 ...
##  $ nox    : num  0.538 0.469 0.469 0.458 0.458 0.458 0.524 0.524 0.524 0.524 ...
##  $ rm     : num  6.58 6.42 7.18 7 7.15 ...
##  $ age    : num  65.2 78.9 61.1 45.8 54.2 58.7 66.6 96.1 100 85.9 ...
##  $ dis    : num  4.09 4.97 4.97 6.06 6.06 ...
##  $ rad    : int  1 2 2 3 3 3 5 5 5 5 ...
##  $ tax    : num  296 242 242 222 222 222 311 311 311 311 ...
##  $ ptratio: num  15.3 17.8 17.8 18.7 18.7 18.7 15.2 15.2 15.2 15.2 ...
##  $ black  : num  397 397 393 395 397 ...
##  $ lstat  : num  4.98 9.14 4.03 2.94 5.33 ...
##  $ medv   : num  24 21.6 34.7 33.4 36.2 28.7 22.9 27.1 16.5 18.9 ...
summary(Boston)
##       crim                zn             indus            chas        
##  Min.   : 0.00632   Min.   :  0.00   Min.   : 0.46   Min.   :0.00000  
##  1st Qu.: 0.08205   1st Qu.:  0.00   1st Qu.: 5.19   1st Qu.:0.00000  
##  Median : 0.25651   Median :  0.00   Median : 9.69   Median :0.00000  
##  Mean   : 3.61352   Mean   : 11.36   Mean   :11.14   Mean   :0.06917  
##  3rd Qu.: 3.67708   3rd Qu.: 12.50   3rd Qu.:18.10   3rd Qu.:0.00000  
##  Max.   :88.97620   Max.   :100.00   Max.   :27.74   Max.   :1.00000  
##       nox               rm             age              dis        
##  Min.   :0.3850   Min.   :3.561   Min.   :  2.90   Min.   : 1.130  
##  1st Qu.:0.4490   1st Qu.:5.886   1st Qu.: 45.02   1st Qu.: 2.100  
##  Median :0.5380   Median :6.208   Median : 77.50   Median : 3.207  
##  Mean   :0.5547   Mean   :6.285   Mean   : 68.57   Mean   : 3.795  
##  3rd Qu.:0.6240   3rd Qu.:6.623   3rd Qu.: 94.08   3rd Qu.: 5.188  
##  Max.   :0.8710   Max.   :8.780   Max.   :100.00   Max.   :12.127  
##       rad              tax           ptratio          black       
##  Min.   : 1.000   Min.   :187.0   Min.   :12.60   Min.   :  0.32  
##  1st Qu.: 4.000   1st Qu.:279.0   1st Qu.:17.40   1st Qu.:375.38  
##  Median : 5.000   Median :330.0   Median :19.05   Median :391.44  
##  Mean   : 9.549   Mean   :408.2   Mean   :18.46   Mean   :356.67  
##  3rd Qu.:24.000   3rd Qu.:666.0   3rd Qu.:20.20   3rd Qu.:396.23  
##  Max.   :24.000   Max.   :711.0   Max.   :22.00   Max.   :396.90  
##      lstat            medv      
##  Min.   : 1.73   Min.   : 5.00  
##  1st Qu.: 6.95   1st Qu.:17.02  
##  Median :11.36   Median :21.20  
##  Mean   :12.65   Mean   :22.53  
##  3rd Qu.:16.95   3rd Qu.:25.00  
##  Max.   :37.97   Max.   :50.00
#Plot matrix of the variables.
pairs(Boston)

There are 506 observations (rows) and 14 variables (columns) in the dataset. The variables are for example crim (per capita crime rate by town), zn (proportion of residential land zoned for lots over 25,000 sq.ft), indus (proportion of non-retail business acres per town), chas (Charles River dummy variable (= 1 if tract bounds river; 0 otherwise)), nox (nitrogen oxides concentration (parts per 10 million)), lstat (lower status of the population (percent)) and medv (median value of owner-occupied homes in $1000s).

For more details of the dataset, please visit https://stat.ethz.ch/R-manual/R-devel/library/MASS/html/Boston.html

library(dplyr)
## 
## Attaching package: 'dplyr'
## The following object is masked from 'package:MASS':
## 
##     select
## The following objects are masked from 'package:stats':
## 
##     filter, lag
## The following objects are masked from 'package:base':
## 
##     intersect, setdiff, setequal, union
library(corrplot)
## corrplot 0.84 loaded
#Calculate the correlation matrix and round it.
cor_matrix<-cor(Boston) %>% round(digits = 2)
#Print the correlation matrix.
cor_matrix
##          crim    zn indus  chas   nox    rm   age   dis   rad   tax ptratio
## crim     1.00 -0.20  0.41 -0.06  0.42 -0.22  0.35 -0.38  0.63  0.58    0.29
## zn      -0.20  1.00 -0.53 -0.04 -0.52  0.31 -0.57  0.66 -0.31 -0.31   -0.39
## indus    0.41 -0.53  1.00  0.06  0.76 -0.39  0.64 -0.71  0.60  0.72    0.38
## chas    -0.06 -0.04  0.06  1.00  0.09  0.09  0.09 -0.10 -0.01 -0.04   -0.12
## nox      0.42 -0.52  0.76  0.09  1.00 -0.30  0.73 -0.77  0.61  0.67    0.19
## rm      -0.22  0.31 -0.39  0.09 -0.30  1.00 -0.24  0.21 -0.21 -0.29   -0.36
## age      0.35 -0.57  0.64  0.09  0.73 -0.24  1.00 -0.75  0.46  0.51    0.26
## dis     -0.38  0.66 -0.71 -0.10 -0.77  0.21 -0.75  1.00 -0.49 -0.53   -0.23
## rad      0.63 -0.31  0.60 -0.01  0.61 -0.21  0.46 -0.49  1.00  0.91    0.46
## tax      0.58 -0.31  0.72 -0.04  0.67 -0.29  0.51 -0.53  0.91  1.00    0.46
## ptratio  0.29 -0.39  0.38 -0.12  0.19 -0.36  0.26 -0.23  0.46  0.46    1.00
## black   -0.39  0.18 -0.36  0.05 -0.38  0.13 -0.27  0.29 -0.44 -0.44   -0.18
## lstat    0.46 -0.41  0.60 -0.05  0.59 -0.61  0.60 -0.50  0.49  0.54    0.37
## medv    -0.39  0.36 -0.48  0.18 -0.43  0.70 -0.38  0.25 -0.38 -0.47   -0.51
##         black lstat  medv
## crim    -0.39  0.46 -0.39
## zn       0.18 -0.41  0.36
## indus   -0.36  0.60 -0.48
## chas     0.05 -0.05  0.18
## nox     -0.38  0.59 -0.43
## rm       0.13 -0.61  0.70
## age     -0.27  0.60 -0.38
## dis      0.29 -0.50  0.25
## rad     -0.44  0.49 -0.38
## tax     -0.44  0.54 -0.47
## ptratio -0.18  0.37 -0.51
## black    1.00 -0.37  0.33
## lstat   -0.37  1.00 -0.74
## medv     0.33 -0.74  1.00
#Visualize the correlation matrix.
corrplot(cor_matrix, method="circle", type="upper", cl.pos="b", tl.pos="d", tl.cex = 0.6)

Positive correlations are displayed in blue and negative correlations in red color. Color intensity and the size of the circle are proportional to the correlation coefficients. There is a high negative correlation between indus (proportion of non-retail business acres per town) and dis (weighted mean of distances to five Boston employment centres), nox (nitrogen oxides concentration (parts per 10 million)) and dis (weighted mean of distances to five Boston employment centres), age (proportion of owner-occupied units built prior to 1940) and dis (weighted mean of distances to five Boston employment centres) and istat (lower status of the population (percent)) and medv (median value of owner-occupied homes in $1000s). There is a high positive correlation between rad (index of accessibility to radial highways) and tax (full-value property-tax rate per $10,000).

Standardize the dataset, create a categorical variable of the crime rate, divide the dataset to train and test sets.

#Center and standardize variables.
boston_scaled <- scale(Boston)
#Summaries of the scaled variables.
summary(boston_scaled)
##       crim                 zn               indus              chas        
##  Min.   :-0.419367   Min.   :-0.48724   Min.   :-1.5563   Min.   :-0.2723  
##  1st Qu.:-0.410563   1st Qu.:-0.48724   1st Qu.:-0.8668   1st Qu.:-0.2723  
##  Median :-0.390280   Median :-0.48724   Median :-0.2109   Median :-0.2723  
##  Mean   : 0.000000   Mean   : 0.00000   Mean   : 0.0000   Mean   : 0.0000  
##  3rd Qu.: 0.007389   3rd Qu.: 0.04872   3rd Qu.: 1.0150   3rd Qu.:-0.2723  
##  Max.   : 9.924110   Max.   : 3.80047   Max.   : 2.4202   Max.   : 3.6648  
##       nox                rm               age               dis         
##  Min.   :-1.4644   Min.   :-3.8764   Min.   :-2.3331   Min.   :-1.2658  
##  1st Qu.:-0.9121   1st Qu.:-0.5681   1st Qu.:-0.8366   1st Qu.:-0.8049  
##  Median :-0.1441   Median :-0.1084   Median : 0.3171   Median :-0.2790  
##  Mean   : 0.0000   Mean   : 0.0000   Mean   : 0.0000   Mean   : 0.0000  
##  3rd Qu.: 0.5981   3rd Qu.: 0.4823   3rd Qu.: 0.9059   3rd Qu.: 0.6617  
##  Max.   : 2.7296   Max.   : 3.5515   Max.   : 1.1164   Max.   : 3.9566  
##       rad               tax             ptratio            black        
##  Min.   :-0.9819   Min.   :-1.3127   Min.   :-2.7047   Min.   :-3.9033  
##  1st Qu.:-0.6373   1st Qu.:-0.7668   1st Qu.:-0.4876   1st Qu.: 0.2049  
##  Median :-0.5225   Median :-0.4642   Median : 0.2746   Median : 0.3808  
##  Mean   : 0.0000   Mean   : 0.0000   Mean   : 0.0000   Mean   : 0.0000  
##  3rd Qu.: 1.6596   3rd Qu.: 1.5294   3rd Qu.: 0.8058   3rd Qu.: 0.4332  
##  Max.   : 1.6596   Max.   : 1.7964   Max.   : 1.6372   Max.   : 0.4406  
##      lstat              medv        
##  Min.   :-1.5296   Min.   :-1.9063  
##  1st Qu.:-0.7986   1st Qu.:-0.5989  
##  Median :-0.1811   Median :-0.1449  
##  Mean   : 0.0000   Mean   : 0.0000  
##  3rd Qu.: 0.6024   3rd Qu.: 0.2683  
##  Max.   : 3.5453   Max.   : 2.9865
#Class of the boston_scaled object.
class(boston_scaled)
## [1] "matrix" "array"
#Change the object to data frame.
boston_scaled <- as.data.frame(boston_scaled)

After scaling the mean is 0 for all the variables which means that all variables are normally distributed.

#Summary of the scaled crime rate.
summary(boston_scaled$crim)
##      Min.   1st Qu.    Median      Mean   3rd Qu.      Max. 
## -0.419367 -0.410563 -0.390280  0.000000  0.007389  9.924110
#Create a quantile vector of crim and print it.
bins <- quantile(boston_scaled$crim)
bins
##           0%          25%          50%          75%         100% 
## -0.419366929 -0.410563278 -0.390280295  0.007389247  9.924109610
#Create a categorical variable 'crime'.
crime <- cut(boston_scaled$crim, breaks = bins, include.lowest = TRUE, labels = c("low", "med_low", "med_high", "high"))
#Look at the table of the new factor crime.
table(crime)
## crime
##      low  med_low med_high     high 
##      127      126      126      127
#Remove original crim from the dataset.
boston_scaled <- dplyr::select(boston_scaled, -crim)
#Add the new categorical value to scaled data.
boston_scaled <- data.frame(boston_scaled, crime)

Divide the dataset to train and test sets, so that 80% of the data belongs to the train set.

#Number of rows in the Boston dataset. 
n <- nrow(boston_scaled)
#Choose randomly 80% of the rows.
ind <- sample(n,  size = n * 0.8)
#Create train set.
train <- boston_scaled[ind,]
#Create test set.
test <- boston_scaled[-ind,]
#Save the correct classes from test data.
correct_classes <- test$crime
#Remove the crime variable from test data.
test <- dplyr::select(test, -crime)

Fit the linear discriminant analysis on the train set.

#Linear discriminant analysis.
lda.fit <- lda(crime ~ ., data = train)
#Print the lda.fit object.
lda.fit
## Call:
## lda(crime ~ ., data = train)
## 
## Prior probabilities of groups:
##       low   med_low  med_high      high 
## 0.2549505 0.2549505 0.2252475 0.2648515 
## 
## Group means:
##                   zn      indus         chas        nox           rm        age
## low       1.01824684 -0.9572354 -0.119431971 -0.8863373  0.507777789 -0.8849674
## med_low  -0.08240509 -0.3295115 -0.004759149 -0.6011270 -0.129299406 -0.4259997
## med_high -0.38169647  0.2130463 -0.012740041  0.4043436 -0.007987863  0.3735181
## high     -0.48724019  1.0149946 -0.125147750  1.0136074 -0.474366363  0.8156853
##                 dis        rad        tax    ptratio       black       lstat
## low       0.9107068 -0.6908519 -0.7272419 -0.4772423  0.38159663 -0.78987402
## med_low   0.4291705 -0.5336346 -0.5102979 -0.0467285  0.32431505 -0.19661420
## med_high -0.3596098 -0.4278305 -0.3024462 -0.2154893  0.06290728  0.08885502
## high     -0.8601874  1.6596029  1.5294129  0.8057784 -0.76473387  0.91732512
##                  medv
## low       0.571329004
## med_low   0.006883708
## med_high  0.073858204
## high     -0.716306617
## 
## Coefficients of linear discriminants:
##                  LD1         LD2         LD3
## zn       0.080677829  0.69406395 -0.92430066
## indus    0.006770828 -0.57250707  0.46978679
## chas     0.046232870  0.07807327  0.17815690
## nox      0.228134482 -0.70909071 -1.42059275
## rm       0.032776869 -0.02550767 -0.16712830
## age      0.229204966 -0.25554348 -0.13490623
## dis     -0.160539828 -0.35804826  0.21001592
## rad      3.597307231  0.83555787  0.41157898
## tax      0.156884078  0.21229003 -0.01802097
## ptratio  0.129919899  0.05617992 -0.29401399
## black   -0.118784530  0.02134071  0.11343129
## lstat    0.120616897 -0.30652047  0.24128032
## medv    -0.020415291 -0.47001620 -0.29044292
## 
## Proportion of trace:
##    LD1    LD2    LD3 
## 0.9594 0.0298 0.0108
#The function for lda biplot arrows.
lda.arrows <- function(x, myscale = 1, arrow_heads = 0.1, color = "orange", tex = 0.75, choices = c(1,2)){
  heads <- coef(x)
  arrows(x0 = 0, y0 = 0, 
         x1 = myscale * heads[,choices[1]], 
         y1 = myscale * heads[,choices[2]], col=color, length = arrow_heads)
  text(myscale * heads[,choices], labels = row.names(heads), 
       cex = tex, col=color, pos=3)
}
#Target classes as numeric.
classes <- as.numeric(train$crime)
#Plot the lda results.
plot(lda.fit, dimen = 2, col = classes, pch = classes)
lda.arrows(lda.fit, myscale = 1)

The best linear separator is variable index of accessibility to radial highways (rad), which has the longest arrow in the picture. The second best separator is difficult to make out, but it looks like it would be nitrogen oxides concentration (parts per 10 million) (nox).

Predicting with the model.

#Predict classes with test data.
lda.pred <- predict(lda.fit, newdata = test)

#Cross tabulate the results.
table(correct = correct_classes, predicted = lda.pred$class)
##           predicted
## correct    low med_low med_high high
##   low       11      12        1    0
##   med_low    5      12        6    0
##   med_high   0      15       18    2
##   high       0       0        1   19

Note that the numbers change every time you run the model. The model predicted the high crime rates well (32/32 were classified correctly). Other categories the model did not predict as well: For medium high crime rates, 17/29 were classified correctly, for medium low crime rates 17/25 were classified correctly, and for low crime rates 9/16 were classified correctly.

total <- c(9+5+2+3+17+5+10+17+2+32)
total
## [1] 102
correct <- c(9+17+17+32)
correct
## [1] 75

Out of a total of 102 observations, 75 observations were classified correctly.

ratio <- c(correct/total)
ratio
## [1] 0.7352941

Accuracy of the model was 74%, which is not that bad but could be better.

Reload Boston dataset.

library(MASS)
data("Boston")
#Center and standardize variables.
boston_scaled <- scale(Boston)
#Change the object to data frame from matrix type.
boston_scaled <- as.data.frame(boston_scaled)
#Calculate the Euclidean distances between observations.
dist_eu <- dist(boston_scaled)
#Look at the summary of the distances.
summary(dist_eu)
##    Min. 1st Qu.  Median    Mean 3rd Qu.    Max. 
##  0.1343  3.4625  4.8241  4.9111  6.1863 14.3970

Run k-means algorithm on the dataset.

#K-means clustering.
km <-kmeans(boston_scaled, centers = 3)
#Plot the Boston dataset with clusters.
pairs(boston_scaled, col = km$cluster)

#Investigate the optimal number of clusters and run the algorithm again.
set.seed(123)
#Determine the number of clusters.
k_max <- 10
#Calculate the total within sum of squares.
twcss <- sapply(1:k_max, function(k){kmeans(boston_scaled, k)$tot.withinss})
#Visualize the results with qplot. Visualize (with qplot) the total WCSS when the number of cluster goes from 1 to 10.
library(ggplot2)
qplot(x = 1:k_max, y = twcss, geom = 'line')

2 clusters seems optimal as the bend (knee) is at 2.

#Run kmeans() again with two clusters.
km <-kmeans(boston_scaled, centers = 2)
#Plot the Boston dataset with clusters.
pairs(boston_scaled, col = km$cluster)

Like observed before, the optimal number of clusters seems to be two.

Super bonus.

model_predictors <- dplyr::select(train, -crime)
#Check the dimensions.
dim(model_predictors)
## [1] 404  13
dim(lda.fit$scaling)
## [1] 13  3
#Matrix multiplication.
matrix_product <- as.matrix(model_predictors) %*% lda.fit$scaling
matrix_product <- as.data.frame(matrix_product)
#Matrix multiplication.
library(plotly)
## 
## Attaching package: 'plotly'
## The following object is masked from 'package:ggplot2':
## 
##     last_plot
## The following object is masked from 'package:MASS':
## 
##     select
## The following object is masked from 'package:stats':
## 
##     filter
## The following object is masked from 'package:graphics':
## 
##     layout
#Create a 3D plot of the columns of the matrix product by typing the code below.
plot_ly(x = matrix_product$LD1, y = matrix_product$LD2, z = matrix_product$LD3, type= 'scatter3d', mode='markers')
## Warning: `arrange_()` is deprecated as of dplyr 0.7.0.
## Please use `arrange()` instead.
## See vignette('programming') for more help
## This warning is displayed once every 8 hours.
## Call `lifecycle::last_warnings()` to see where this warning was generated.
#Add argument color as an argument in the plot_ly() function.
plot_ly(x = matrix_product$LD1, y = matrix_product$LD2, z = matrix_product$LD3, type= 'scatter3d', mode='markers', color = train$crime)

Draw another 3D plot where the color is defined by the clusters of the k-means.

#Make a k-means with 4 clusters to compare the methods.
km3D <-kmeans(boston_scaled, centers = 4)
plot_ly(x = matrix_product$LD1, y = matrix_product$LD2, z = matrix_product$LD3, type= 'scatter3d', mode='markers', color = km3D$cluster[ind])

Medium high crime rates seems to bee better defined than cluster 1. Cluster 2 is a bit better defined (not so much intermingling) than low crime rate in the first picture. Cluster 3 is quite similar with high crime rates, even though cluster 3 is a bit more better defined. Cluster 4 is more defined than medium low crime rates in the first picture.